Permutations Fundamentals

The multiplication principle from the previous lessons has a compact formula when you are arranging $r$ objects chosen from a larger group of $n$ distinct objects: the number of such arrangements is written $P(n,r)$. This lesson formalizes that formula and practices recognizing when a problem calls for it, including problems that combine several independent groups.

What you will learn

• Apply the permutation formula P(n,r)
• Count arrangements of a subset of a larger group
• Use the multiplication principle to build permutation counts

Keep paper and a pencil nearby. Try each problem before opening a hint, and mark sections reviewed as you work through them.

The P(n,r) formula

When arranging $r$ positions chosen from $n$ distinct objects, with no repeats, the first position has $n$ choices, the second has $n-1$, the third has $n-2$, and so on, stopping after $r$ factors. This count is written $P(n,r)$, and it equals $n\times(n-1)\times(n-2)\times\cdots\times(n-r+1)$, a product of exactly $r$ consecutive decreasing whole numbers starting at $n$.

Using factorials, $P(n,r)=\dfrac{n!}{(n-r)!}$, because expanding $n!$ only down to $(n-r)!$ and cancelling the shared $(n-r)!$ leaves exactly the $r$ factors from $n$ down to $n-r+1$. Both the direct product form and the factorial form give the same value; use whichever is faster for the numbers in front of you.

Recognizing when to apply P(n,r)

Look for two signals together: a larger pool of $n$ distinct objects, and a smaller number $r$ of positions or roles being filled from that pool, where each object can be used at most once and the assignment to positions matters. Electing a president, vice president, and secretary from a club of candidates is a textbook $P(n,r)$ situation: there are more candidates than roles, each candidate can hold at most one role, and swapping which candidate holds which role produces a different outcome.

It is easy to misidentify $r$: count the number of distinct roles or positions being filled, not the number of candidates available. A common error is using $n$ where $r$ belongs, which produces a full arrangement of everyone instead of the requested partial arrangement.

Combining several independent groups

Many contest problems arrange more than one group in the same scenario — for example, choosing 2 letters from one set and 2 digits from another to build a code. When the groups are independent (the choice in one group does not affect what is available in the other), compute the number of arrangements within each group separately, then multiply the group totals together, exactly as the multiplication principle allows for any sequence of independent choices.

Before multiplying, double check that each group's count already accounts for its own "no repeats" or ordering condition; a common mistake is multiplying two correct-looking numbers where one of them secretly used the wrong pool size or forgot the ordering condition entirely.

See the reasoning unfold

Pause after each step and predict what comes next.

Common mistakes to catch

• Using the pool size $n$ as the number of factors instead of the number of positions $r$ actually being filled.
• Forgetting to multiply the counts of two or more independent groups in a combined problem.
• Applying $P(n,r)$ to a situation where order does not actually matter, overcounting the true number of outcomes.

Try with support

Guided practice

Choose a strategy, take a first step, and use the hints when you need direction.

Put it into practice

Independent practice

Work on your own first. If you reveal a solution, work through it and mark it reviewed.

Bring it together

• $P(n,r)=n\times(n-1)\times\cdots\times(n-r+1)=\dfrac{n!}{(n-r)!}$ counts arrangements of $r$ positions filled from a pool of $n$.
• Identify $n$ as the pool size and $r$ as the number of positions actually being filled.
• For independent groups, compute each group's count separately and multiply the totals.